POJ 2361 Tic Tac Toe 模拟

最新推荐文章于 2021-01-29 13:55:24 发布

kdwycz

最新推荐文章于 2021-01-29 13:55:24 发布

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本文介绍了一个TicTacToe游戏的有效性判断算法。通过分析输入的棋盘状态，判断该状态是否能由合法的游戏过程得出。算法检查了棋盘上X和O的数量差异，并验证了获胜条件的一致性。

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Tic Tac Toe

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1992		Accepted: 750

Description

Tic Tac Toe is a child's game played on a 3 by 3 grid. One player, X, starts by placing an X at an unoccupied grid position. Then the other player, O, places an O at an unoccupied grid position. Play alternates between X and O until the grid is filled or one player's symbols occupy an entire line (vertical, horizontal, or diagonal) in the grid.
We will denote the initial empty Tic Tac Toe grid with nine dots. Whenever X or O plays we fill in an X or an O in the appropriate position. The example below illustrates each grid configuration from the beginning to the end of a game in which X wins.

...  X..  X.O  X.O  X.O  X.O  X.O  X.O

...  ...  ...  ...  .O.  .O.  OO.  OO.

...  ...  ...  ..X  ..X  X.X  X.X  XXX

Your job is to read a grid and to determine whether or not it could possibly be part of a valid Tic Tac Toe game. That is, is there a series of plays that can yield this grid somewhere between the start and end of the game?

Input

The first line of input contains N, the number of test cases. 4N-1 lines follow, specifying N grid configurations separated by empty lines.

Output

For each case print "yes" or "no" on a line by itself, indicating whether or not the configuration could be part of a Tic Tac Toe game.

Sample Input

2
X.O
OO.
XXX

O.X
XX.
OOO

Sample Output

yes
no

Source

Waterloo local 2002.09.21

#include <iostream>
using namespace std;
int main()
{
	string a[4];
	int f,n,wo,wx,no,nx,i,j;
	cin>>n;
	while (n--){
		wo=wx=no=nx=0; f=1;
		for (i=0;i<3;++i)
			cin>>a[i];
		for (i=0;i<3;++i)
			for (j=0;j<3;++j){
				if (a[i][j]=='O')
					no++;
				if (a[i][j]=='X')
					nx++;
			}
				
	
		if (a[0][0]==a[0][1] && a[0][1]==a[0][2]){
			if (a[0][1]=='O') wo=1;
			if (a[0][1]=='X') wx=1;
		}
		if (a[1][0]==a[1][1] && a[1][1]==a[1][2]){
			if (a[1][1]=='O') wo=1;
			if (a[1][1]=='X') wx=1;
		}
		if (a[2][0]==a[2][1] && a[2][1]==a[2][2]){
			if (a[2][1]=='O') wo=1;
			if (a[2][1]=='X') wx=1;
		}
		if (a[0][0]==a[1][0] && a[1][0]==a[2][0]){
			if (a[1][0]=='O') wo=1;
			if (a[1][0]=='X') wx=1;
		}
		if (a[0][1]==a[1][1] && a[1][1]==a[2][1]){
			if (a[1][1]=='O') wo=1;
			if (a[1][1]=='X') wx=1;
		}
		if (a[0][2]==a[1][2] && a[1][2]==a[2][2]){
			if (a[1][2]=='O') wo=1;
			if (a[1][2]=='X') wx=1;
		}
		if (a[0][0]==a[1][1] && a[1][1]==a[2][2]){
			if (a[1][1]=='O') wo=1;
			if (a[1][1]=='X') wx=1;
		}
		if (a[0][2]==a[1][1] && a[1][1]==a[2][0]){
			if (a[1][1]=='O') wo=1;
			if (a[1][1]=='X') wx=1;
		}
		
		if (no>nx) f=0;
		if (nx-no>=2) f=0;
		if (wo && wx) f=0;
		if (wo && no!=nx) f=0;
		if (wx && no==nx) f=0;
		
		if (f)
			cout<<"yes\n";
		else
			cout<<"no\n";
		
	}
	return 0;
}

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